时间序列摘要

Question

我试图对时间滞后为 1 的数字求和。即，我想通过添加特定组中天数仅相差一天的值的频率来总结行。我使用了滞后函数来获取差异，但不知道如何从这里开始。

df <- df %>% 
  group_by(group) %>% 
  mutate(diff = dt - lag(dt))

df[!is.na(df$diff) & df$diff > 1,]$diff <- NA

例如：

 group     dt           freq  diff  
 groupA    2016-03-21    1     NA    
 groupA    2016-03-22    1     1     
 groupA    2016-03-23    1     1     
 groupA    2016-03-26    2     NA     
 groupA    2016-03-28    1     NA     
 groupA    2016-03-29    3     1     
 groupA    2016-03-30    3     1     
 groupA    2016-03-31    5     1     
 groupB    2016-04-01    1     NA      
 groupB    2016-04-02    2     1 

我需要把它分成：

group    dt         freq  diff  duration     
groupA  2016-03-21    1     NA    3 (1 + 1 + 1)     
groupA  2016-03-22    1     1         
groupA  2016-03-23    1     1         
groupA  2016-03-26    2     NA    2     
groupA  2016-03-28    1     NA    12(1 + 3 + 3 + 5)     
groupA  2016-03-29    3     1         
groupA  2016-03-30    3     1         
groupA  2016-03-31    5     1         
groupB  2016-04-01    1     NA    3(1 + 2)     
groupB  2016-04-02    2     1 

也提到了this，但累积不起作用，因为我不考虑跳跃超过一天。循环自定义函数是唯一的方法吗？

Answer 1

您可以使用这种方法更轻松地做到这一点（以less.than 1 天的差异对行进行分组）；这将创建一个帮助列 gap 稍后将用于对同一组中连续几天的 freq 求和：

library(dplyr)

df %>% 
    mutate(gap = cumsum(!c(TRUE, diff(as.Date(df$dt)) == 1)))  %>% 
    group_by(gap, group) %>% 
    mutate(duration = sum(freq, na.rm=TRUE)) %>% 
    ungroup %>% select(-gap) %>% as.data.frame

#            group         dt freq duration
#        1  groupA 2016-03-21    1        3
#        2  groupA 2016-03-22    1        3
#        3  groupA 2016-03-23    1        3
#        4  groupA 2016-03-26    2        2
#        5  groupA 2016-03-28    1       12
#        6  groupA 2016-03-29    3       12
#        7  groupA 2016-03-30    3       12
#        8  groupA 2016-03-31    5       12
#        9  groupB 2016-04-01    1        3
#        10 groupB 2016-04-02    2        3

Answer 2

这是使用dplyr::lead 的tidyverse 解决方案：

library(tidyverse);
df %>%
    mutate(dt = as.POSIXct(dt)) %>%
    group_by(group) %>%
    mutate(
        diff = pmin(c(1, diff(dt)), c(1, diff(lead(dt))), na.rm = T),
        id = cumsum(c(TRUE, diff(diff) != 0) | diff > 1)) %>%
    group_by(group, id) %>%
    mutate(duration = sum(freq)) %>%
    ungroup() %>%
    select(-diff, -id)
## A tibble: 10 x 4
#   group  dt                   freq duration
#   <fct>  <dttm>              <int>    <int>
# 1 groupA 2016-03-21 00:00:00     1        3
# 2 groupA 2016-03-22 00:00:00     1        3
# 3 groupA 2016-03-23 00:00:00     1        3
# 4 groupA 2016-03-26 00:00:00     2        2
# 5 groupA 2016-03-28 00:00:00     1       12
# 6 groupA 2016-03-29 00:00:00     3       12
# 7 groupA 2016-03-30 00:00:00     3       12
# 8 groupA 2016-03-31 00:00:00     5       12
# 9 groupB 2016-04-01 00:00:00     1        3
#10 groupB 2016-04-02 00:00:00     2        3

解释：diff 选择前后日期之间的最小差值。然后我们寻找diff 的变化，并创建一个新的分组向量id，我们通过它计算汇总度量sum(freq)。

---

样本数据

df <- read.table(text =
    " group     dt           freq  diff
 groupA    2016-03-21    1     NA
 groupA    2016-03-22    1     1
 groupA    2016-03-23    1     1
 groupA    2016-03-26    2     NA
 groupA    2016-03-28    1     NA
 groupA    2016-03-29    3     1
 groupA    2016-03-30    3     1
 groupA    2016-03-31    5     1
 groupB    2016-04-01    1     NA
 groupB    2016-04-02    2     1 ", header = T)

---

更新

第二个例子：

# Sample data
df <- read.table(text =
" group     dt           freq  diff
groupA    2016-03-21    1     NA
groupA    2016-03-22    1     1
groupA    2016-03-23    1     1
groupA    2016-03-26    2     NA
groupA    2016-03-28    1     NA
groupA    2016-04-01    3     1
groupA    2016-04-02    3     1
groupA    2016-04-03    5     1
groupB    2016-04-01    1     NA
groupB    2016-04-02    2     1 ", header = T)

df %>%
    mutate(dt = as.POSIXct(dt)) %>%
    group_by(group) %>%
    mutate(
        diff = pmin(c(1, diff(dt)), c(1, diff(lead(dt))), na.rm = T),
        id = cumsum(c(TRUE, diff(diff) != 0) | diff > 1)) %>%
    group_by(group, id) %>%
    mutate(duration = sum(freq)) %>%
    ungroup() %>%
    select(-diff, -id);
## A tibble: 10 x 4
#   group  dt                   freq duration
#   <fct>  <dttm>              <int>    <int>
# 1 groupA 2016-03-21 00:00:00     1        3
# 2 groupA 2016-03-22 00:00:00     1        3
# 3 groupA 2016-03-23 00:00:00     1        3
# 4 groupA 2016-03-26 00:00:00     2        2
# 5 groupA 2016-03-28 00:00:00     1        1
# 6 groupA 2016-04-01 00:00:00     3       11
# 7 groupA 2016-04-02 00:00:00     3       11
# 8 groupA 2016-04-03 00:00:00     5       11
# 9 groupB 2016-04-01 00:00:00     1        3
#10 groupB 2016-04-02 00:00:00     2        3